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Prepared and Edited by: HEMLATA AGGARWAL
Maths-II
Unit-C Laplace Transforms and its Applications
Definition: Let )(tf be a function of t defined for all 0t . The Laplace transform of )(tf ,
denoted by )(tfL , is defined as
0
)()( dttfetfL st
provided that the integral exists, ‘s’ is a parameter which may be real or complex.
)(tfL is clearly a function of s and is briefly written as )(sf i.e )(tfL = )(sf .
Linearity property: If 21 ,cc are constants and f, g are functions of t,
then
)()()()( 2121 tgLctfLctgctfcL
Piecewise continuous function: A function F(t) is said to be piecewise continuous on a closed interval
bta , if it is defined on that interval and is such that the interval can be subdivided into a finite
number of intervals in each of which F(t) is continuous and has finite right and left hand limits.
Functions of Exponential order: A function F(t) is said to be of exponential order as t if there
exists a positive constant(real) M, a number and a finite number 0t such that atMetF )( or
MtFe at )( , for all 0t .
If a function F(t) is of exponential order , it is also of , .
Existence of Laplace Transform: If F(t) is a function which is piecewise continuous on every finite
interval in the range 0t and satisfies atMetF )( for all 0t and for some constants a and M, then
the Laplace transform of F(t) exists for all p > a.
Prepared and Edited by: HEMLATA AGGARWAL
Laplace transform of some elementary functions:
(1) 0,1
1 ss
L (2) ,!1n
n
s
ntL where n is a +ve integer
(3) asas
eL at ,1
(4) 0,sin22
sas
aatL
(5) 0,cos22
sas
satL (6) as
as
aatL ,sinh
22
(7) asas
satL ,cosh
22
First shifting theorem (multiplication by ate ):
If )(tfL = )(sf , then )( )( asftfeL at .
By definition,0
)(
0
)()()( dttfedttfeetfeL tasatstat
0
)( dttfe pt , where p = s - a
)()( asfpf
Similarly, )()( asftfeL at .
Applying this theorem to the elementary functions, we get the following useful results
Prepared and Edited by: HEMLATA AGGARWAL
(1) 1)(
!n
nat
as
nteL (2)
22)(sin
bas
bbteL at
(3) 22)(
cosbas
asbteL at (4)
22)(sinh
bas
bbteL at
(5) 22)(
coshbas
asbteL at
Change of scale property:
If )(tfL = )(sf , then )( 1
)(a
sf
aatfL .
Proof: 0
)()( dtatfeatfL st du
dt du,adt u,atput a
0
).(a
duufe a
su
af
aatfL
s where(u)du e
1)(
0
u-
a
sf
af
aufL
a
1)(
1)(
1
Example: Find the Laplace transform: (1) )5sin35cos2(3 tte t (2) btat cos.cosh
Solution: (1) )5sin(3)5cos(2)5sin35cos2( 333 teLteLtteL ttt
346
92
5)3(
53
5)3(
32
22222 ss
s
ss
s
Prepared and Edited by: HEMLATA AGGARWAL
(2) L(coshat.cosbt) bteeL atat cos).(2
1bteLbteL atat cos
2
1cos
2
1
By first shifting theorem
assass bs
s
bs
s2222 2
1
2
12222 )(2
1
)(2
1
bas
as
bas
as
Inverse Laplace transform:
Definition: If )()( sftfL , then f(t) is called the inverse Laplace transform of )(sf and is denoted
by
)()(1 tfsfL
Here 1L denotes the inverse Laplace transform.
Inverse Laplace transform of some elementary functions:
(1) 111
sL
(2) ,)1(
1 11
n
t
sL
n
n where n is a positive
integer
(3))!1()(
1 11
n
te
asL
nat
n (4) at
as
sL cos
22
1
(5) atas
sL cosh
22
1 (6) btebas
asL at cos
)( 22
1
Prepared and Edited by: HEMLATA AGGARWAL
(7) )cos(sin2
1
)(
1322
1 atatataas
L (8) ateas
L11
(9) ataas
L sin11
22
1 (10) ataas
L sinh11
22
1
(11) btebbas
L at sin1
)(
122
1
(12) attaas
sL sin
2
1
)( 222
1
Laplace Transform of Derivative: If )(tf be continuous and )()( sftfL , then
0)(lim),0()()( tfeprovidedfssftfL st
t
Proof: dttfetfL st
0
)()( dttfsetfe stst
0
0)()(
dttfesftfe stst
t)()0()(lim
0
)0()()()0(0 fsfstfsLf
General Form: If f(t) and its first (n-1) derivative be continuous, then
)0()0()0()0()()( 1321 nnnnnn ffsfsfssfstfL
Prepared and Edited by: HEMLATA AGGARWAL
Laplace Transform of Integrals: If )()( sftfL , then
duufsfs
Lorsfs
duufL
tt
0
1
0
)()(1
)(1
)(
Proof: Let 0)0( and)()( then ,)()(
0
tftduuft
t
. )0()()( sstL
)(1
)()()()()( sfs
tLtsLsfsstfL
tt
duufsfs
Lorsfs
duufL0
1
0
)()(1
)(1
)(
Hence, if t
duufsfs
LthentfsfL0
11 )()(1
)()(
Multiplication by nt :
If ),()( sftfL then ,)()1()( sfds
dtftL
n
nnn
where n = 1, 2, … .
Proof: We prove the theorem by induction
)()()()(
0
sfdttfesftfL st
Differentiating both sides w.r.t s (using Leibnitz’s rule for differentiation under the integral sign), we
have
)()(
)()(
0
0
sfds
ddttfe
sor
sfds
ddttfe
ds
d
st
st
Prepared and Edited by: HEMLATA AGGARWAL
)()(
)()(
0
0
sfds
ddtttfeor
sfds
ddttfteor
st
st
),()( sfds
dttfL which confirm the truth of the theorem for n = 1
Now assume the theorem to be true for n = m, so that
)()1()(
)()1()(
0
sfds
ddttfteor
sfds
dtftL
m
mmmst
m
mmm
Differentiating both sides, we have
)()1()(1
1
0
sfds
ddttfte
ds
dm
mmmst
)()1()(1
1
0
sfds
ddttfte
s m
mmmst
)()1()(1
1
0
sfds
ddttftte
m
mmmst
)()1()(1
11
0
1 sfds
ddttfte
m
mmmst
)()1()(1
111 sf
ds
dtftL
m
mmm
which shows that the theorem is true for n = m+1. Hence by mathematical induction, the theorem is
true for all positive integer n.
Corollary: If .)()1()(),()( 11 tftsfds
dLthentfsfL nn
n
n
Prepared and Edited by: HEMLATA AGGARWAL
.)()(1 ttfsfds
dL
Division by t:
If ),()( sftfL then dssftft
L
s
)()(1
, provided integral exists.
Proof: We have .)()(
0
dttfesf st Integrating both sides w.r.t s from s to , we have
s
st
s
dsdttfedssf
0
)()(
Since s and t are independent variable, changing the order of integration on the right hand side, we have
0
)()( dttfdsedssf
s
st
s
.)(1)(
)(00
tft
Ldtt
tfedttf
t
e st
s
st
CONVOLUTION THEOREM
If )()(1 tfsfL and )()(1 tgsgL then .*)()()()(0
1 gfduutgufsgsfL
t
(f g is called the convolution of f and g).
Proof: Let
t
duutguft0
)()()(
dtduutgufedtduutgufetL
t
st
t
st
0 00 0
)()()()()(
On changing the order of integration, we get
Prepared and Edited by: HEMLATA AGGARWAL
0
)()()(u
st dtduutgufetL dudtutgeufe
u
utssu
0
)( )()(
dudvvgeufeu
svsu
0
)()( , on putting t - u = v
duufesgdusgufe susu
00
)()()()( )().()().( sgsfsfsg
t
duutguftsgsfL0
1 .)()()()()(
Example: Use the convolution theorem to evaluate
(1) 222
21
)( as
sL (2)
22
21
)4(s
sL
Solution: (1) 222
21
)( as
sL
By the Convolution Theorem, we have
2222222
21 .
)( as
s
as
s
as
sL
t
duutaau0
)(coscos
t
duauatauatau0
)sinsincos(coscos
t t
duauauatauduat0 0
2 sincossincoscos
t t
duauatduauat0 0
2sinsin2
1)2cos1(cos
2
1
tt
aua
ataua
uat00
2cos2
1sin
2
12sin
2
1cos
2
1
)2cos1(sin4
12sin
2
1cos
2
1atat
aat
atat
Prepared and Edited by: HEMLATA AGGARWAL
)2cossincos2(sin4
1sin
4
1cos
2
1atatatat
aat
aatt
)2sin((sin4
1cos
2
1atatat
aatt )sincos(
2
1atatat
a
(2) 22
21
)4(s
sL
We have ts
sL 2cos
)2( 22
1 . By the Convolution theorem
2222222
21
2.
2)2( s
s
s
s
s
sL
t
duutu0
)(2cos2cos
t
duututu0
)2sin2sin2cos2(cos2cos
t t
duuutudut0 0
2 2sin2cos2sin2cos2cos
t t
duutduut0 0
4sin2sin2
1)4cos1(2cos
2
1
tt
utuut00
4cos4
12sin
2
14sin
4
12cos
2
1
)4cos1(2sin8
14sin
4
12cos
2
1ttttt
)4cos2sin2cos4(sin8
12sin
8
12cos
2
1ttttttt ))24sin(2(sin
8
12cos
2
1ttttt
)2sin2cos2(4
1ttt
Example: Find the Laplace transform of .sin2 att
Solution: 22
sinas
aatL
Prepared and Edited by: HEMLATA AGGARWAL
222
222 )1(sin
as
a
ds
dattL
322
22
222
2
)(
)3(2
)(
2)1(
as
asa
as
as
ds
d
Example: Find the Laplace transform of t
e t )1(.
Solution: 1
11)()1()1(
sseLLeL tt
s
t
dssst
eL
1
11)1(
sss )1log(log
s
s
s
s
s
s
1log
11
1log
1log
Example: Evaluate tdtte t sin0
2
Solution: dtttetdtte tt )sin(sin0
2
0
2 ,where s = 2
)sin( ttL
25
4
)12(
4
)1(
2
1
1)1(
22222 s
s
sds
d.
Application to Differential equations: Laplace transform can be used to solve ordinary as well
as partial differential equations. We shall apply this method to solve only ordinary linear
differential equations with constant coefficients. The advantage of this method is that it yields
the particular solution directly without the necessity of first finding the general solution and then
evaluating the arbitrary constants.
Algorithm:
Take Laplace transforms of both sides of the given differential equation, using initial conditions. This gives an algebraic equation.
Solve the algebraic equation to get y in terms of s.
Prepared and Edited by: HEMLATA AGGARWAL
Take inverse Laplace transform of both sides. This gives y as a function of t which is the desired solution.
Remember: 1 2 1( ) ( ) (0) '(0) ........ (0).n n n n nL f t s f s s f s f f
Example: Solve the equation
3 2 2
3 2 22 2 0, where 1, 2, 2at 0.
d y d y dy dy d yy y t
dt dt dt dt dt
Solution: The given equation is ''' 2 '' ' 2 0y y y y
Taking the Laplace transform of both sides, we get
3 2 2(0) '(0) ''(0) 2 (0) '(0) (0) 2 0s y s y sy y s y sy y sy y y (A)
Using the given conditions (0) 1, '(0) 2, ''(0) 2,y y y equation (A) reduces to
3 2 22 2 2 2 2 4 1 0s s s y s s s
3 2 2
2 2
3 2
2 2 4 5
4 5 4 5 5 1 1(By PartialFractions)
2 2 ( 1)( 1)( 2) 3( 1) 1 3( 2)
s s s y s s
s s s sy
s s s s s s s s s
Taking the inverse Laplace transform of both sides, we get
Prepared and Edited by: HEMLATA AGGARWAL
1 1 1
2
5 1 1 1 1
3 1 1 3 2
1(5 ) , which is the requiredsolution.
3
t t t
y L L Ls s s
y e e e
Application to Integral Equations
An equation in which an unknown function occurs inside an integral is called an integral equation. Thus,
an equation of the form
( ) ( ) ( ) ( , )
b
a
Y t F t Y u K u t du (1)
in which F(t) and K(u, t) are known functions and Y(t) is the unknown function is an integral equation.
Here a and b are either constants or functions of t. The function K(u, t) is often called the kernel of the
integral equation. If a and b are constants, equation (1) is called Fredholm integral equation. If a is a
constant while b = t, it is called a Volterra integral equation.
A special integral equation of convolution type is
0
( ) ( ) ( ) ( )
t
Y t F t Y u G t u du
The Laplace transform is an excellent tool for solving such integral equations of convolution type.
Example: Solve the integral equation
2
0
( ) ( )sin( )
t
y t t y u t u du
Solution: Given equation is 2( ) ( )siny t t y t t
Taking Laplace transform and using convolution theorem, we have
Prepared and Edited by: HEMLATA AGGARWAL
3 3 2
2 3
2
2 3
3 5
2 2 1( ) ( ) . sin ( ).
1
1 2( ) 1
1
2( ).
1
2 2( )
y s L y t L t y ss s s
y ss s
sy s
s s
y ss s
Taking inverse Laplace transform, we get
1 1
3 5
42
1 1( ) 2 2
( ) .12
y t L Ls s
ty t t
Laplace Transform of some other useful functions
1. Unit Step Function (or Heaviside’s Unit Function):
The unit step function is defined as
0,,1
,0)( awhere
atfor
atforatu .
As a particular case,
0,1
0,0)(
tfor
tfortu
Prepared and Edited by: HEMLATA AGGARWAL
The product
atfortf
atforatutf
),(
,0)().(
The function )()( atuatf represents the graph of f(t) shifted through a distance ‘a’ to the
right.
Now Laplace Transform of Unit Step Function is
0
)()( dtatueatuL st as
a
st
a
st
a
st ess
edtedte
101.0
0
In particular s
tuL1
)(
Second Shifting Theorem:
If ),()( sftfL then 0
)().()().( dtatuatfeatuatfL st
0 0
)( ,)()( duufedtatfe ausst where u = t - a
)()(0
sfeduufee assuas
Corollary 1: )().()(1 atuatfsfeL as
Corollary 2: If a = 0, .)()()()( tfLsftutfL
Example: Use Laplace transform method to solve
Prepared and Edited by: HEMLATA AGGARWAL
2
22 with 2, 1 0td x dx dx
x e x at tdt dt dt
Solution: Taking the Laplace transform of both sides, we get
1
1)0(2)0()0(
___2
sxxxsxsxxs
Using the given conditions we get
1
67252
1
1)12(
2_2
s
sss
sxss
323
2_
)1(
1
)1(
3
1
2
)1(
672
ssss
ssx (By partial fractions)
3
1
2
11
)1(
1
)1(
13
1
12
sL
sL
sLx
22
.2
132
!2
.
!1
32 tetee
tetee ttt
ttt
2. Unit Impulse Function (or Dirac-delta Function)
In mechanics, we come across problems where a very large force acts for a very small time. In the study
of bending of beams, we have point loads which is equivalent to large pressure acting over a very small
area. To deal with such problems, we introduce the unit impulse function or Dirac-delta function. Unit
impulse function is considered as the limiting form of the function
atfor
atafor
atfor
at
,0
,1
,0
)(
Integrating this function
Prepared and Edited by: HEMLATA AGGARWAL
1)(11
)(
0
a
a
aadtdtat
As 0 , the function )( at tends to be infinite at x = a and zero elsewhere, with the
characteristic property that its integral across t=a is unity. If )( at represents a force acting for
short duration at time t = a, then the integral
1)(lim0
a
a
dtat
represents unit impulse at t = a. Hence the limiting form of (t-a) as 0 is expressed as unit
impulse function denoted by )( at .
Thus the unit impulse function )( at is defined as follows :
atfor
atforat
,0
,)( , such that 1)(
0
dtat
Laplace Transform of Unit Impulse Function
If )(tf be a function of t continuous at t=a, then
dttfdtattf
a
a
1).()()(
0
)(1
)()( cfcfaa , where a < c < a
(by Mean-value Theorem for integrals)
As 0 , we get )()()(0
afdtattf
Corollary 1: sast edtateatL0
)()(
Corollary 2: 1)( 0setL
Prepared and Edited by: HEMLATA AGGARWAL
3. Laplace Transform of Periodic Functions
If )(tf is a periodic function with period T, i.e. f(t + T) = f(t), then
.)(1
1)(
0
dttfee
tfL
T
st
sT
Proof: dttfedttfedttfedttfetfL
T
T
st
T
T
st
T
stst )()()(.)()(
3
2
2
00
Putting t = u, t = u+T, t = u+2T,…… in the successive integrals
duTufeduTufeduufetfL
T
Tus
T
Tus
T
su )2()()()(0
)2(
0
)(
0
Since )2()()( TufTufuf , we have
duufeeduufeeduufetfL
T
susT
T
susT
T
su )()()()(0
)2
00
dttfee
duufeeetfL
T
st
sT
T
susTsT )(1
1)()1()(
00
2 .
Example: Find the Laplace transform of
)1()1( 2 tut
Solution: Comparing )1()1( 2 tut with )()( atuatf , we have
a = 1 and 2)( ttf
3
2)()(
stfLsf
)()1()1( 2 sfetutL s (by Second Shifting theorem)
Prepared and Edited by: HEMLATA AGGARWAL
3
2
s
e s
Example: Fnd the Laplace transform of the triangular wave function of period 2c given by
.2,2
0,)(
ctctc
ctttf
Solution: )(tfL dttcetdtee
dttfee
c
c
st
c
st
cs
c
st
cs
2
0
2
2
0
2)2(.
1
1)(
1
1
c
c
ststc
stst
cs s
e
s
etc
s
e
s
et
e
2
2
0
22).1().2(.1.
1
1
22
2
222
1
1
1
s
ce
s
ce
s
e
ss
e
s
ce
e
cscscscscs
cs
cs
cs
cscs
cscscs
cs e
e
see
e
ss
ee
e 1
1.
1
)1)(1(
)1(.
121
1
12
2
2
2
2
2
tanh11
222
22
2
cs
see
ee
s cscs
cscs
